$ A = \left[\begin{array}{rrr}-1 & 5 & 5 \\ 4 & 5 & 4\end{array}\right]$ $ F = \left[\begin{array}{rr}3 & 3 \\ -2 & -1 \\ 5 & 5\end{array}\right]$ What is $ A F$ ?
Solution: Because $ A$ has dimensions $(2\times3)$ and $ F$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ A F = \left[\begin{array}{rrr}{-1} & {5} & {5} \\ {4} & {5} & {4}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{3} \\ {-2} & \color{#DF0030}{-1} \\ {5} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{3}+{5}\cdot{-2}+{5}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{3}+{5}\cdot{-2}+{5}\cdot{5} & ? \\ {4}\cdot{3}+{5}\cdot{-2}+{4}\cdot{5} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{3}+{5}\cdot{-2}+{5}\cdot{5} & {-1}\cdot\color{#DF0030}{3}+{5}\cdot\color{#DF0030}{-1}+{5}\cdot\color{#DF0030}{5} \\ {4}\cdot{3}+{5}\cdot{-2}+{4}\cdot{5} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{3}+{5}\cdot{-2}+{5}\cdot{5} & {-1}\cdot\color{#DF0030}{3}+{5}\cdot\color{#DF0030}{-1}+{5}\cdot\color{#DF0030}{5} \\ {4}\cdot{3}+{5}\cdot{-2}+{4}\cdot{5} & {4}\cdot\color{#DF0030}{3}+{5}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}12 & 17 \\ 22 & 27\end{array}\right] $